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Thermodynamics Basics

Heat energy and temperature change — how much energy it takes to heat a material, and how much a component grows when it gets hot. Essential for boiler design, pipe systems, and precision machining.

Q = mcΔT ΔL = αLΔT Specific heat capacity Linear expansion
2Topics
8Examples
4Calculators
12Practice Qs
🔥Specific Heat CapacityQ = mcΔT
🌡️Thermal ExpansionΔL = αLΔT
⚙️Applications4 workshop calculators
🎮Game12 randomised questions
📝Practice12 graded questions
Why this matters in engineeringReal world

Heating a tank of water to process temperature requires knowing the specific heat capacity — too small a heater and the process takes too long. Thermal expansion determines the clearance gaps in engine components, railway tracks and bridge expansion joints. Get it wrong and parts seize or buckle.

🔥

Specific Heat Capacity

The energy needed to raise 1 kg of a material by 1 °C. Different materials absorb heat at different rates.

📜 Key rules

Heat energy:  Q = m × c × ΔT  — mass (kg) × specific heat capacity (J/kg°C) × temperature rise (°C).
Unit of heat energy: joules (J). Large quantities in kJ (1000 J) or MJ (1 000 000 J).
Specific heat capacity of water: c = 4200 J/kg°C. Steel: ≈ 500 J/kg°C. Aluminium: ≈ 900 J/kg°C.
ΔT is the temperature change — final temperature minus initial temperature.
🔥Specific Heat Capacity CalculatorChoose what to find, enter values, press Calculate
Choose what to find, enter the other values, then press Calculate.

📚 Key vocabulary

Specific heat capacity (c)
the energy needed to raise 1 kg of a substance by 1 °C, in J/kg°C.
Heat energy (Q)
the thermal energy transferred to or from a substance, measured in joules.
Temperature change (ΔT)
the difference between final and initial temperature, in °C or K — the same numerical value.
Thermal mass
the ability of a material to absorb and store heat; high mass and high c means slow heating.
Kilojoule (kJ)
1000 joules — used for large heat energy values such as heating water.
Megajoule (MJ)
1 000 000 joules — used for very large energy quantities such as industrial process heating.
👁️ Visual aid — heat energy flow: flame heats a 5 kg water tank from 20 °C to 100 °C, Q = m × c × ΔT
Water m = 5 kg 🔥 T₁ = 20 °C T₂ = 100 °C ΔT = 80 °C Q = m × c × ΔT = 1 680 000 J
✍️Worked example 1 — heating water in a boiler
QuestionA boiler heats 5 kg of water from 20 °C to 100 °C. c = 4200 J/kg°C. Calculate the heat energy needed.
Step 1ΔT = 100 − 20 = 80 °C
FormulaQ = m × c × ΔT
SubstituteQ = 5 × 4200 × 80
Step 25 × 4200 = 21 000
Step 321 000 × 80 = 1 680 000
AnswerQ = 1 680 000 J = 1680 kJ
✍️Worked example 2 — heating an aluminium casting
QuestionAn aluminium casting of mass 3 kg is heated from 25 °C to 205 °C. c = 900 J/kg°C. Find Q in kJ.
Step 1ΔT = 205 − 25 = 180 °C
SubstituteQ = 3 × 900 × 180
Step 23 × 900 = 2700
Step 32700 × 180 = 486 000
Convert486 000 ÷ 1000 = 486
AnswerQ = 486 kJ
✍️Worked example 3 — finding mass from Q
Question630 000 J heats a steel component by 60 °C. c = 500 J/kg°C. Find the mass.
Formulam = Q ÷ (c × ΔT)
Denominator500 × 60 = 30 000
Substitutem = 630 000 ÷ 30 000
Divide630 000 ÷ 30 000 = 21
Answerm = 21 kg
✍️Worked example 4 — finding temperature rise
Question504 000 J is supplied to 4 kg of copper (c = 390 J/kg°C). Find the temperature rise.
FormulaΔT = Q ÷ (m × c)
Denominator4 × 390 = 1560
SubstituteΔT = 504 000 ÷ 1560
Divide504 000 ÷ 1560 = 323.08
AnswerΔT ≈ 323 °C
🌡️

Thermal Expansion

Materials expand when heated. Linear expansion calculates the increase in length of a rod, pipe or beam for a given temperature rise.

📜 Key rules

Linear expansion:  ΔL = α × L × ΔT  — coefficient of expansion (/°C) × original length (m) × temperature change (°C).
Coefficient of linear expansion α: steel ≈ 12 × 10−&sup6; /°C, aluminium ≈ 23 × 10−&sup6; /°C, copper ≈ 17 × 10−&sup6; /°C.
The result ΔL is very small — usually in millimetres. Multiply by 1000 to convert metres to mm.
Final length:  Lf = L + ΔL. Always add the expansion to the original length.
🌡️Thermal Expansion CalculatorChoose what to find, enter values, press Calculate
Choose what to find, enter the other values, then press Calculate.

📚 Key vocabulary

Thermal expansion
the increase in length, area or volume of a material when its temperature rises.
Linear expansion (ΔL)
the increase in one dimension (length) of a solid when heated.
Coefficient of linear expansion (α)
the fractional change in length per degree Celsius — unique to each material.
Expansion joint
a deliberate gap in a structure (bridge, pipe, rail) that allows thermal expansion without buckling.
Clearance fit
a small designed gap between mating parts so they can still move freely at operating temperature.
Interference fit
a joint where a shaft is cooled (or a hub heated) so expansion locks the parts together on return to room temperature.
👁️ Visual aid — linear expansion: a 10 m steel rail (α = 12×10−&sup6;) heated 50 °C expands by 6 mm
L = 10 m (α = 12×10−&sup6;) ΔL = 6 mm 🌡️ +50 °C ΔL = 12×10−&sup6; × 10 × 50 = 0.006 m
✍️Worked example 5 — expansion of a steel rail
QuestionA steel rail is 10 m long at 10 °C. The temperature rises to 60 °C. α = 12 × 10−&sup6; /°C. Find the expansion.
Step 1ΔT = 60 − 10 = 50 °C
FormulaΔL = α × L × ΔT
SubstituteΔL = 12 × 10−&sup6; × 10 × 50
Step 212 × 10−&sup6; × 10 = 1.2 × 10−&sup4;
Step 31.2 × 10−&sup4; × 50 = 6 × 10−³ m
AnswerΔL = 0.006 m = 6 mm
✍️Worked example 6 — expansion of an aluminium pipe
QuestionAn aluminium pipe is 4 m long and heats by 75 °C. α = 23 × 10−&sup6; /°C. Calculate the expansion in mm.
SubstituteΔL = 23 × 10−&sup6; × 4 × 75
Step 223 × 10−&sup6; × 4 = 92 × 10−&sup6;
Step 392 × 10−&sup6; × 75 = 6900 × 10−&sup6; = 0.0069 m
AnswerΔL = 0.0069 m = 6.9 mm
✍️Worked example 7 — finding temperature rise from measured expansion
QuestionA copper rod 2 m long expands by 0.0034 m. α = 17 × 10−&sup6; /°C. Find the temperature rise.
FormulaΔT = ΔL ÷ (α × L)
Denominator17 × 10−&sup6; × 2 = 34 × 10−&sup6;
SubstituteΔT = 0.0034 ÷ 34 × 10−&sup6;
Divide0.0034 ÷ 0.000034 = 100
AnswerΔT = 100 °C
✍️Worked example 8 — final length of a steel shaft
QuestionA steel shaft is 0.5 m long at 20 °C and heats to 120 °C. α = 12 × 10−&sup6; /°C. Find its final length.
Step 1ΔT = 120 − 20 = 100 °C
ExpansionΔL = 12 × 10−&sup6; × 0.5 × 100 = 0.0006 m
Final lengthLf = 0.5 + 0.0006 = 0.5006 m
AnswerLf = 0.5006 m (0.6 mm longer)
⚙️

Engineering Applications

Four thermal calculators for process heating, pipe design and machining scenarios.

🔥 Process tank heating

Calculate the energy needed to heat a tank of liquid to working temperature — helps size the heater element.

Q = m × c × ΔT
🧮 Calculator
Enter mass, c and temperature rise, then press Calculate.

🌡️ Pipeline expansion gap

Find how much an installed pipe will expand so the correct expansion joint gap can be set.

ΔL = α × L × ΔT
🧮 Calculator
Enter α, length and temperature rise, then press Calculate.

⏱️ Heater power & time

Given the required heat energy and the heater power, find the time the heater must run.

t = Q ÷ P
🧮 Calculator
Enter mass, c, ΔT and heater power, then press Calculate.

🔩 Interference fit — shaft heating

Find the temperature rise needed to expand a steel shaft enough to fit into a housing bore.

ΔT = ΔL ÷ (α × L)
🧮 Calculator
Enter required expansion, diameter and α, then press Calculate.
🎮

Thermal Challenge

12 randomised questions per round — new numbers every time you play.

🔥 Quick-fire round

Question 1 / 12 Score: 0 Streak: 0 🔥
Press Start to begin!
📝

Practice Questions

12 graded questions. Attempt on paper first, then reveal the full worked solution.

c values: water 4200, steel 500, aluminium 900, copper 390 J/kg°C. α values: steel 12, aluminium 23, copper 17 (×10−&sup6;/°C). Show formula, substitution and all steps.
Question 1 · SHC1 mark
Calculate the heat energy needed to raise 2 kg of water by 50 °C. c = 4200 J/kg°C.
Q = m × c × ΔT = 2 × 4200 × 50 = 420 000 J = 420 kJ
Question 2 · SHC2 marks
A steel component of mass 8 kg is heated from 20 °C to 320 °C. c = 500 J/kg°C. Calculate Q in kJ.
Q = m × c × ΔT
ΔT = 320 − 20 = 300 °C
Q = 8 × 500 × 300 = 1 200 000 J = 1200 kJ
Question 3 · SHC2 marks
756 000 J heats a copper block through 90 °C. c = 390 J/kg°C. Find the mass of the block.
m = Q ÷ (c × ΔT)
m = 756 000 ÷ (390 × 90)
390 × 90 = 35 100
756 000 ÷ 35 100 = 21.54 kg ≈ 21.5 kg
Question 4 · SHC2 marks
A 6 kg aluminium casting absorbs 972 000 J. c = 900 J/kg°C. Find the temperature rise.
ΔT = Q ÷ (m × c)
ΔT = 972 000 ÷ (6 × 900)
6 × 900 = 5400
972 000 ÷ 5400 = 180 °C
Question 5 · SHC2 marks
A workshop heater supplies 3 kW for 10 minutes to heat a steel block. How much heat energy is delivered, and what temperature rise does this produce in a 5 kg block? c = 500 J/kg°C.
Q = P × t = 3000 × 600 = 1 800 000 J
ΔT = Q ÷ (m × c) = 1 800 000 ÷ (5 × 500) = 1 800 000 ÷ 2500 = 720 °C
Question 6 · Expansion1 mark
A steel rod is 5 m long and heats by 40 °C. α = 12 × 10−&sup6; /°C. Find the expansion in mm.
ΔL = α × L × ΔT = 12e−6 × 5 × 40 = 0.0024 m = 2.4 mm
Question 7 · Expansion2 marks
An aluminium beam is 8 m long at 15 °C. At 65 °C, find its new length. α = 23 × 10−&sup6; /°C.
ΔT = 65 − 15 = 50 °C
ΔL = 23e−6 × 8 × 50 = 0.0092 m
Lf = 8 + 0.0092 = 8.0092 m
Question 8 · Expansion2 marks
A copper pipe 3 m long expands by 1.53 mm when heated. α = 17 × 10−&sup6; /°C. Find the temperature rise.
ΔT = ΔL ÷ (α × L)
= 0.00153 ÷ (17e−6 × 3)
= 0.00153 ÷ 0.000051 = 30 °C
Question 9 · Expansion2 marks
A railway track must allow 9.6 mm of expansion per rail. Each rail is 20 m long, α = 12 × 10−&sup6; /°C. Find the maximum temperature rise the gap allows.
ΔT = ΔL ÷ (α × L)
= 0.0096 ÷ (12e−6 × 20)
= 0.0096 ÷ 0.00024 = 40 °C
Question 10 · Expansion2 marks
A steel shaft is 0.08 m long and needs to expand 0.00096 m to fit a housing. α = 12 × 10−&sup6; /°C. How much must it be heated?
ΔT = ΔL ÷ (α × L)
= 0.00096 ÷ (12e−6 × 0.08)
= 0.00096 ÷ 0.00000096 = 1000 °C
Question 11 · Combined3 marks
A 10 kg steel tank holds water. The water (mass 200 kg) and tank are heated from 15 °C to 75 °C. Find the total heat energy. cwater = 4200, csteel = 500 J/kg°C.
ΔT = 75 − 15 = 60 °C
Qwater = 200 × 4200 × 60 = 50 400 000 J
Qsteel = 10 × 500 × 60 = 300 000 J
Qtotal = 50 400 000 + 300 000 = 50 700 000 J = 50 700 kJ
Question 12 · Combined3 marks
A 3 kW boiler heats 30 kg of water from 20 °C to 80 °C (c = 4200 J/kg°C). (a) Find Q. (b) Find the time in minutes.
Q = 30 × 4200 × 60 = 7 560 000 J
t = Q ÷ P = 7 560 000 ÷ 3000 = 2520 s
2520 ÷ 60 = 42 minutes
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